Question

Bài 1 Tìm \(x\)

a)\(\left(\dfrac{1}{2}\times x-3\right)\times\left(-\dfrac{1}{3}+x\right)=0\)

b)\(\dfrac{1}{2}\times x^2-\dfrac{1}{5}\times x=0\)

c)\(\dfrac{1`}{4}\times x=\dfrac{1}{16}\times x^2\)

d)\(9\times x^2=1\)

e)\(\left(x-5\right)^2=4\)

Answer 1

a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)

Answer 2

d) \(9x^2=1\)

\(\Leftrightarrow x^2=1:9\)

\(\Leftrightarrow x^2=\dfrac{1}{9}\)

\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

Answer 3

e) \(\left(x-5\right)^2=4\)

\(\Leftrightarrow\left(x-5\right)^2=2^2\)

\(\Leftrightarrow x-5=2\)

\(\Leftrightarrow x=2+5\)

\(\Leftrightarrow x=7\)

Answer 4

a) Để \((\dfrac{1}{2}\times x-3)\times(-\dfrac{1}{3}+x)=0\) thì

\(\left[{}\begin{matrix}\dfrac{1}{2}\times x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}\times x=3\\x=\dfrac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\div\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy x=6 hoặc x=1/3

d) Ta có: \(9\times x^2=1\)

=> \(x^2=\dfrac{1}{9}\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)

Vậy x=1/3 hoặc x=-1/3

e)Ta có: \(\left(x-5\right)^2=4\)

=> \(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

Vậy x=7 hoặc x=3

Answer 5

Các bạn ơi có ai muốn biết chuyện tình đồi thông 2 mộ ở Đà Lạt ko?